PHYSICS 315
Solution to problem set 6:

4.3-1:
(a)
Substituting $a=4.05\times 10^{-10}$ m into formula for conduction electron density gives

$$n= \frac{3\times 4}{a^3}=1.806\times 10^{-29} \;{\rm m^{-3}}$$

(using the fact that Al is trivalent and that the fcc lattice has 4 atoms/unit cube) we find for the conductivity, with $m_e=9.105\times 10^{-31}$ kg, $e=-1.602\times 10^{-19}$ C, $\tau=0.8\times 10^{-14}$ s$^{-1}$

$$\sigma=\frac{ne^2\tau}{m_e}=4.07\times 10^7 \;{\rm ohm^{-1}m^{-1}}$$

b: With $\epsilon_0=8.8542\times 10^{-12} {\rm Fm^{-1}}$ the plasma frequency is

$$\Omega_{pl}=\sqrt{\frac{ne^2}{\epsilon_0m}}=2.4\times 10^{16} {\rm rad\; s^{-1}}$$

c: The wavelength for light of the same frequency is

$$\lambda=\frac{2\pi c}{\Omega_{pl}}=78.6\; {\rm nm}$$

4.3-2 We have for the wavevector (with the approximation that $\omega\tau>>1$

$$q=\frac{\omega}{c}\sqrt{\epsilon_r}=\frac{\sqrt{\omega^2-\Omega_{pl}^2}}{c}$$

we find

$$\frac{dq}{d\omega}=\frac{\omega}{c\sqrt{\omega^2-\Omega_{pl}^2}}$$

giving for the group velovity

$$v_g=\frac{d\omega}{dq}=c\sqrt{1-\frac{\Omega_{pl}^2}{\omega^2}}<c$$

c The index of refraction is

$$n=\sqrt{1-\frac{1}{4}}$$

giving

$$\theta_c=60^o$$

for the critical angle.
4.3-3
When the intensity is reduced to 1/2

$$z=\frac{c\ln(2)}{2\omega k}$$

where $k$ is the imaginary part of the relative dielectric constant. Substituting for $\Omega{pl}$ from 4.3-1b and $\omega=2\pi 10^7 {\rm rad\; sec^{-1}}$ and the value of $\tau$ into

$$\epsilon_r=1-\frac{\Omega_pl^2}{\omega(\omega+\frac{i}{\tau})}$$

gives

$$k=1.91\times 10^5$$

and $z=8.6\;\mu$m. For $\omega=2\pi 10^{14}{\rm rad\; sec^{-1}}$ I find $k=37.6$ and $z=4.4$ nm.